Another AC anomaly?
in [Logic]
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From: Virgil on 18 Dec 2009 21:16 In article <b8a6551e-b171-46c9-8e3b-3de00eb350d8(a)v25g2000yqk.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 17 Dez., 22:47, Quaestor <quaes...(a)qqq.com> wrote: > > > > Please prove your assertion by showing which path I did not produce. > > > But prove it by knowing only all nodes of all paths. > > > > Until you give some explicit way of telling which paths you do produce, > > ther is no way of telling which ones you miss. > > I produce every path. No you don't. In particular, you do not produce any path which contains both infinitely many left branchings and infinitely many right branchings, but there are most common kind in any ACTUAL complete infinite binary tree. > There is no node in the complete binary tree > that is not member of an infinitude of paths. That is still true for demonstrably incomplete sets of paths, as well as many sets of finite sub-paths, so is totally irrelevant to the issue. > > > > > Information exceeding the infinite sequence of nodes of each path is > > > not acceptable in mathematics. > > > > I do not accept your fiats on what is or is not acceptable in > > mathematics, > > In mathematics, numbers are defined by digits. WRONG! Numerals can be defined by digits but numbers can be defined in a number of ways which do not require digits at all. For example, the number pi is standardly defined with no reference to digits at all. And so is the number e. > Additional information > is not part of mathematics. If you do not accept that, then there is > no common basis for discussion. There seems to be no basis for anyone with a reasonable knowledge of mathematics for to have a discussion of mathematics with someone like WM who ignores so much mathematics in order to support his delusions about mathematics. > > My claim is: The set of numbers which can unambiguously be identified > by digit sequences is countable. But in asddition, you claim that these are all of the numbers there are, even though such numbers as e and pi have no known sequence of digits which unambiguously identify them, so cannot be a part of WM's odd version of mathematics. > > > > > > Which path do you see in the tree that has not that property? > > > > In my tree, none. But in there are sets of nodes satisfying my condition > > of pathhood which do not occur in your tree. > > Pure assertion. Give an example, please. In my tree, each path represents a real number between 1 and 2 by inserting a binary radix point after the root node and representing each node of form 2*n + i, i in {0,1}, by its i value. Then, for example, pi/2 corresponds to a path in my tree but not in yours.
From: Virgil on 18 Dec 2009 21:22 In article <3678078d-7bde-4457-896a-2d55ee158088(a)y24g2000yqb.googlegroups.com>, WM <mueckenh(a)rz.fh-augsburg.de> wrote: > On 17 Dez., 22:52, Quaestor <quaes...(a)qqq.com> wrote: > > > > > > For paths and initial segments to contain and to be is the same. In > > > > > fact: > > > > > {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} > > > > > and also > > > > > {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} = {1, 2, 3, ...}. > > > > > > As usual you confuse to be an element and to be a party of... > > > > > I am concerned with numbers. Your absurd distinctions are irrelevant. > > > > They are neither absurd nor irrelevant to mathematics. > > Ok, then you have a different idea of mathematics. In common > mathematics, numbers can be defined by digit sequences. Not all of them. Negatives require a negative sign , which is not a digit. Decimal numbers reqire a decimal point, which is not a digit. In fact, the only numbers that can be defined by digits alone are the non-negative integers > > > > > > Either N exists or not. If yes, then it is the union of all natural > > > numbers as well as the union of all initial segments of the ordered > > > set of all natural numbers as well as the union of all these and > > > itself, because > > > > > {1, 2, 3, ...} U {1, 2, 3, ...} = {1, 2, 3, ...} > > > > > If you see that this equation is true, then you acknowledge my > > > argument. > > > > On the contrary, that equation is true > > Fine, Virgil, that is one big step to comprehension. But however hard WM tries, he has not what is needed to attain it.
From: WM on 19 Dec 2009 08:10 On 18 Dez., 15:12, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > Aha, you are clearly a mindreader. Well, as far as I know mindreading > > > is not part of mathematics. Anyhow, I can think of numbers larger than > > > that path. > > > But that is completely irrelevant. I am able to think about a set that > > > contains all natural numbers, you apparently are not. > > > > How do you know that without confirming it by thinking the last too? > > Why need I to think about a last one (which there isn't) to be able to think > about a set that contains all natural numbers? Apparently you have some > knowledge about how my mind works that I do not have. Yes. A very convincing and often required proof of completenes of a linear set is to know the last element. T talk about all in case there is no last is silly. > > > > > > Right, but there is no finite initial segment that contains them all. > > > > > > > > That is pure opinion, believd by the holy bible (Dominus regnabit in > > > > aeternum et ultra. [2. Buch Moses: Exodus 15 Vers 18]) or forced upon > > > > us by the men-made axiom of infinity. > > > > > > Sorry, I have no knowledge of the bible. But live without that axiom when > > > you can't stomach it. And do not attack mathematicians who live with that > > > axiom. > > > > To live with that axiom does not create uncountability. See the proof > > here: > >http://groups.google.com/group/sci.logic/browse_frm/thread/46fa18c8bb.... > > Where is the proof there? I see only you writing a bit of nonsense and two > rebuttals. One of the rebuttals has meanwhile been changed. Peter Webb recognized: It is true that you cannot show pi as a finite decimal, but you can't show 1/3 as a finite decimal either. Just what I said. > > > > > The tree contains all paths that can be constructed by nodes, using > > > > the axiom of infinity. Which one would be missing? > > > > > > The infinite paths because you stated a priori that your tree did not > > > contain infinite paths. So it is impossible to construct in your tree > > > infinite paths by the axiom of infinity. > > > > The axiom of infinity establishes the set N from finite numbers. > > It establishes the *existence* of a set N of finite numbers. What else should be established? > > > It establishes the infinite paths as well in my tree from finite > > paths. > > No. That is impossible because you stated that the paths were finite. > What it *does* establish is the extistence of a set P of finite paths. It is rather silly to argue about the uncountability of the set of paths. Only minds completely disformed by set theory could try to defend the obviously false position that there were uncountably many paths. But "10 Questions" will give you the answer why there are not uncountably many paths. There are no infinite decimal expansions of real numbers. There are not due paths in the tree. It is as impossible to express any real number by an infinite decimal expansion as it is impossible to express 0 by a unit fraction. The rest will be explained in "10 Questions". Therefore I will stop with this topic here. Regards, WM
From: WM on 19 Dec 2009 08:17 On 18 Dez., 15:19, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <2cd3c08a-a072-4ef5-bc0b-f0aaa9126...(a)a21g2000yqc.googlegroups..com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 17 Dez., 15:08, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <6a57309a-a136-430c-a718-e38518c65...(a)q16g2000vbc.googlegroups= > > .com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > > > On 16 Dez., 03:44, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > > > > ({1} U {1, 2} U {1, 2, 3} U ... = ({1} U {1, 2} U {1, 2, 3} U > > > > > > U {1, 2, 3, ...} > > > > > > > > That is a matter of taste. > > > > > > No, it is a matter of convention. In mathematics > > > a, b, c, ..., z > > > means a, b, c, continue this way until you reach z. But starting > > > {1}, {1, 2}, {1, 2, 3} > > > and going on you never reach > > > {1, 2, 3, ...} > > > > That is true. Therefore it does not exist. > > That you can not get there step by step does not mean that it does not > exist. That you cannot get step by step to 1/0 does not mean that it does not exist? > > > However, see Cantor, > > collected works, p 445: > > 0, 1, 2, 3, ... w_0, w_0 + 1, ..., gamma, ..., > > He seems to reach far more. > > Right, he uses a convention that is no longer used. Wrong, it is used presently, for instance by myself. > > > > > > A sequence of paths is not a path. > > > > > > > > But a union of paths is. > > > > > > No. Suppose we have the paths 0.000 and 0.100, what is their union? And > > > is it a path? > > > > No. But the union contains two paths. > > Wrong. If we look at the paths as sets, they are sets of nodes. Their > union is a set of nodes, not a set of paths. And as a set of nodes we > can form from them seven different paths. Wrong. The nodes of two paths give exactly two paths. But as I already said, it is so silly to argue about the uncountability of the paths in the tree that I will stop it here. > > > > Let every finite path of every infinite path be mapped on the elements > > of omega. That was simple. > > By your statements infinite paths do not exist. But pray give such a > mapping. Until now you have only asserted that such a mapping exists > without showing that. Do you accept the mapping from omega on SUM{k = 1 to n} 3*10^-k yields the infinite decimal expansion of 1/3? Regards, WM
From: WM on 19 Dec 2009 08:21
On 18 Dez., 15:22, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <dd852490-e622-4bc4-b858-dfcc3b142...(a)l13g2000yqb.googlegroups..com> WM <mueck...(a)rz.fh-augsburg.de> writes: > > On 17 Dez., 20:49, Virgil <Vir...(a)home.esc> wrote: > ... > > > WM <mueck...(a)rz.fh-augsburg.de> wrote: > > > > > No. Suppose we have the paths 0.000 and 0.100, what is their union? > > > > > And is it a path? > > > > > > > No. But the union contains two paths. And an infinite union of that > > > > kind may contain the path 0.111... > > > > > > So according to WM a union of sets may contain an object not contained > > > in any of the sets being unioned. > > > > For paths and initial segments to contain and to be is the same. > > Eh? paths contain nodes and initial segments contain numbers. > But be aware that you use the word 'contains' with two different meanings > at different times. For paths we have: Paths contain initial segments and paths are initial segments. > > > For paths and initial segments to contain and to be is the same. In > > fact: > > {1} U {1, 2} U {1, 2, 3} U ... = {1, 2, 3, ...} > > and also > > {1} U {1, 2} U {1, 2, 3} U ... U {1, 2, 3, ...} = {1, 2, 3, ...}. > > In the mathematical sense the union contains numbers, not sets. The union contains subsets. > > > This is so according to set theory. Of course it is rubbish. > > Well, if you want to use terminology in a different meaning than standard, > of course. To contain as a subset is not correct in English? Regards, WM |